2007 Alabama ARML TST Problems/Problem 14
Contents
Problem
Find the sum of the real roots of
Solution
Step 1: using the symmetry
Let
The polynomial is symmetric. Each symmetric polynomial has the following property: for , is a root if and only if is.
To see why this is true, assume that we have a such that . We can now divide both sides by . We get:
But the left hand side is just . Thus also .
Step 2: multiplying three quadratic terms
Our polynomial has six complex roots: , , , , , and . It can be expressed as:
Let , , and . We can then write:
We can now compare the coefficients of , and in this expression. (Comparing other coefficients is not necessary due to symmetry.) We get the following equations:
Simplifying, we get:
Using Vieta's formulas, this means that , and are the roots of the polynomial .
Step 3: Finding a root of q
Obviously, is increasing for , and . Thus has exactly one positive root. We can easily find that . Thus WLOG .
(Note: Remembering that and , we now have and . From the first equation, let . Substituting into the second one, we get . This obviously has no real solution, thus and are not real. But we don't need this observation.)
Step 4: Partially dividing p
We now know that is a factor of . We can partially divide to get:
Step 5: Handling the remaining part of p
The polynomial has no real roots.
To prove this, one can for example write:
Step 6: Summary
We found that , where has no real roots. Thus all real roots of are the two real roots of , and their sum is obviously .
Note
roots of p computed by Octave: -4.48006 + 14.30270i -4.48006 - 14.30270i 8.88748 + 0.00000i 0.11252 + 0.00000i -0.01994 + 0.06367i -0.01994 - 0.06367i
See also
2007 Alabama ARML TST (Problems) | ||
Preceded by: Problem 13 |
Followed by: Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 |